Question

(UFMG) Os números a, b e c são as raízes da equação x³ + x - 1 =0. Nessas condições, calculem o valor de log (1/a + 1/b + 1/c)

Answer 1

$\displaystyle \sf x^3+x-1=0 \\\\ ra{\'i}}zes : a,b , c \\\\ \log\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\ ?$

Vamos usar as relações de girard :

$\displaystyle \sf x^3+0x^2+x-1 = 0 \\\\ a+b+c = \frac{-0}{1} = 0 \\\\ ab+bc+ac = \frac{1}{1} = 1 \\\\ abc = \frac{-(-1)}{1} = 1 \\\\\ temos: \\\\ \log\left(\frac{1}{ a}+\frac{1}{b}+\frac{1}{c}\right) \\\\\\ Tirando\ o \ MMC:\\\\\ \log\left(\frac{1}{a/bc}+\frac{1}{b/ac}+\frac{1}{c/ab}\right) \\\\\\ \log\left(\frac{ab+bc+ac}{abc}\right) \\\\\\\ \log\left(\frac{1}{1}\right) \\\\\\ \log(1) = 0 \\\\ Portanto : \\\\ \boxed{\sf \log\left(\frac{1}{ a}+\frac{1}{b}+\frac{1}{c}\right) = 0 }\checkmark$